Answer:
The fraction of its volume inside liquid  is increased . 
Explanation:
According to principle pf floatation , an object floats on the surface of water 
when the weight of  liquid displaced by it becomes equal to weight of the object . weight of the liquid depends upon the density of the liquid . 
In the second case , when the body is dipped into liquid of lesser density , in order to balance the weight of body , more volume of liquid will be displaced so that weight of displaced liquid becomes equal to object's weight . So the body floats with greater depth inside liquid . The fraction of its volume inside liquid  is increased . 
 
        
             
        
        
        
Answer:
1.-E=1000N/C to the LEFT
2.-The electric field inside a conductor in electrostatic state is always zero (conductor proprieties). 
3.-The voltmeter read 0V as differential voltage between two points from the conductor
Explanation:
1.The electric field inside the conductor must be zero (conductor proprieties). Then the charges create a electric field equal an opposite to the external electric field. In other words E=1000N/C to the LEFT
2. The electric field inside a conductor in electrostatic state is always zero. As shown in the figure the electric field induced by the charges in the sphere surface cancelled the EXTERN electric field.
3.If the Electric field inside the conductor is zero, that means that the Voltage in the hole conductor is constant (conductor proprieties). In other words the the voltmeter read 0v as differential voltage between two points from the conductor.
 
        
             
        
        
        
Answer:
C
Explanation:
a series circuit would be an odd choice to power a battery or light a lamp when a direct would be much more efficient, and it's not converting types of energy, so C is the best possible answer 
 
        
             
        
        
        
Answer:
ΔV=0.484mV
Explanation:
The potential difference across the end of conductor that obeys Ohms law:
ΔV=IR
Where I is current
R is resistance
The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A
R=(pL)/A
Given data
Length L=3.87 cm =0.0387m
Diameter d=2.11 cm =0.0211 m
Current I=165 A
Resistivity of aluminum p=2.65×10⁻⁸ ohms
So
ΔV=IR

ΔV=0.484mV